Chem 4501 Introduction to Thermodynamics, 3 Credits Kinetics, and Statistical Mechanics Fall Semester 2017

Homework Problem Set Number 10—Solutions 1. McQuarrie and Simon, 10-4. Paraphrase: Apply Euler’s theorem to internal energy to derive a fundamental equation.

Given that U depends on 3 extensive variables, i.e.,

U =U S,V,n( ) we can apply Euler’s theorem to obtain

U S,V,n( ) = S ∂U∂S

"

#$

%

&'V ,n

+V ∂U∂V"

#$

%

&'S,n

+ n ∂U∂n

"

#$

%

&'S,V

and we know from the natural variables of U that the partial derivative quantities are themselves simple thermodynamic quantities

U S,V,n( ) = S T( )+V −P( )+ n µ( )

= TS −PV +G

we may rearrange to

G =U −TS +PV = A+PV = H −TS which is the definition of the Gibbs free energy.

2. McQuarrie and Simon, 10-16 (ignore the reference to problem 10-14, which is not needed). Paraphrase: From the lever rule compute the relative amounts of liquid and vapor phases if a mixture has an overall 50:50 composition and the liquid and vapor phases have mole fractions for one component of 0.38 and 0.57, respectively.

The lever rule tells us that the total number of moles of liquid relative to vapor can be computed from

€

nliqnvap

=y2 − xaxa − x2

=0.57 − 0.500.50 − 0.38

=0.070.12

= 0.58

Or, put differently, there is about a 37:63 ration of liquid to vapor.

3. McQuarrie and Simon, 10-17 and 10-18. Paraphrase: Derive the analytic form of the pressure/mole-fraction curves for the liquid and vapor phases of an ideal binary solution as a function of the pure component vapor pressures. Discuss the relationship between the vapor pressures of the two pure components and the relative enrichment of the liquid vs. vapor phases in one of them.

For an ideal solution, the vapor pressure is the mole-fraction weighted average of the pure component vapor pressures. Thus,

€

Ptot = x1P1* + x2P2

* If we take advantage of the relationship that x1 = (1 – x2), then we have

€

Ptot = 1− x2( )P1* + x2P2

*

= P2* − P1

*( )x2 + P1*

which indicates that the total vapor pressure is a linear function of x2 with slope P2* – P1* and intercept P1*. In the gas-phase, Dalton’s law says that the total pressure will derive from the contributions of all individual partial pressures, which are weighted equally on a per-mole basis, so the mole fraction of component 2 will be

€

y2 =x2P2

*

Ptot

or, expressed in terms of x2

€

x2 =y2Ptot

P2*

Inserting this expression into the total vapor pressure expression derived above provides

€

Ptot = P2* − P1

*( ) y2Ptot

P2*

+ P1*

which may be rearranged to give

€

Ptot =P1

*P2*

P2* − y2 P2

* − P1*( )

and a curved plot of total pressure will be observed as a function of y2. Returning to the expression for y2 above, and substituting for Ptot as derived further above, we have

€

y2 =x2P2

*

P2* − P1

*( )x2 + P1*

The goal is to express the relationship of y2 to x2 in terms of the pure component vapor pressures. So, first we can divide both sides by x2

€

y2x2

=P2

*

P2* − P1

*( )x2 + P1*

Since we want to understand how the ratio on the right-hand side varies as a function of the ratio of the pure component vapor pressures, we should divide numerator and denominator on the r.h.s. by P1*.

€

y2x2

=P2

* / P1*

P2* / P1

*( ) −1# $ %

& ' ( x2 +1

As we’re interested in whether the ratio on the l.h.s. is less than or greater than 1, we should subtract 1 from both sides, arriving at

€

y2x2

−1 =P2

* / P1*

P2* / P1

*( ) −1# $ %

& ' ( x2 +1

−P2

* / P1*( ) −1#

$ % & ' ( x2 +1

P2* / P1

*( ) −1# $ %

& ' ( x2 +1

=P2

* / P1*( ) 1− x2( ) − 1− x2( )

P2* / P1

*( ) −1# $ %

& ' ( x2 +1

=P2

* / P1*( ) −1#

$ % & ' (

1− x2( )

P2* / P1

*( ) −1# $ %

& ' ( x2 +1

We are finally in a position to evaluate the behavior of the r.h.s. for the alternative cases of P1* > P2* and vice versa. For the former case, the quantity in square brackets on the r.h.s. (in both the numerator and denominator) is negative. In the denominator, since the magnitude of the value in brackets is less than unity (think about it) and the value of x2 is a positive number less than unity, the net sign of the denominator is positive, so the net sign of the r.h.s. is negative, and thus y2 < x2. A similar argument establishes the converse. This proves the intuitive result that the vapor phase above an ideal binary solution will be enriched in the component with the higher vapor pressure.

4. McQuarrie and Simon, 10-24. Paraphrase: Prove that the partial molar volume of a component in an ideal binary solution is equal to the molar volume of the pure component liquid.

The partial molar volume is defined from the total volume, generally, as

€

Vtot = n1V 1 + n2V 2

In addition, we know from Maxwell relations that

€

Vtot =∂G∂P#

$ %

&

' ( T ,n1,n2

In the case of an ideal binary solution, G is equal to the sum of the chemical potentials of the components, written as

€

G = n1µ1 + n2µ2

= n1 µ1* + RT ln x1( ) + n2 µ2

* + RT ln x2( )

So, if we differentiate with respect to P holding the other three variables constant, we have

€

∂G∂P#

$ %

&

' ( T ,n1,n2

= n1∂µ1

*

∂P

#

$ % %

&

' ( ( T ,n1,n2

+ 0 + n2∂µ2

*

∂P

#

$ % %

&

' ( ( T ,n1,n2

+ 0

= n1V 1* + n2V 2

*

Comparing the two approaches to Vtot then makes evident that the partial molar volume in solution is the same as that for the pure component liquid.

5. McQuarrie and Simon, 10-32. Paraphrase: Show that if the chemical potential of a solute varies according to Henry’s law as its mole fraction concentration goes to zero then the chemical potential of the solvent follows Raoult’s law as its mole fraction correspondingly goes to unity.

We start with the definition of the chemical potential of the solute (component 2)

€

µ2 = µ2* + RT ln P2

P2*

= µ2* + RT ln P2 − RT ln P2

*

Under Henry’s law conditions, P2 approaches kH,2x2 as x2 à 0, so under those conditions we may write

€

µ2 = µ2* + RT ln kH,2x2 − RT ln P2

*

= µ2* − RT ln P2

* + RT ln kH,2 + RT ln x2

= µ2o + RT ln kH,2 + RT ln x2

where in the final line we’ve used the relationship between the pure liquid standard state (*) and the 1-bar ideal gas standard state (o) to simplify notation. We can now take the differential of each side, noting that the first two terms on the r.h.s. are simply constants, to determine

€

dµ2 = RTdx2x2

From the Gibbs-Duhem equation, we have

€

x1dµ1 = −x2dµ2 Solving for dµ1 after substituting for dµ2 using our above derivation provides

€

dµ1 = −x2x1

dµ2

= −x2x1

RTdx2x2

= RTdx1x1

where in the last line we’ve used dx2 = –dx1 (from mass balance). Using dummy integration variables in this equation and integrating from x1 = a (a value just slightly less than 1) to x1 = 1 gives

€

dµ1"µ1" x1=a( )

µ1" x1=1( )∫ = RT

dx1"

x1"a1∫

µ1* −µ1 x1 = a( ) = RT ln1− RT ln a

= −RT ln a

Or, rearranged, and replacing a with x1

€

µ1 = µ1* + RT ln x1

which is the desired result, noting that if x2 à 0 then x1 à 1.

6. McQuarrie and Simon, 10-37. Paraphrase: Derive the dependence of the molar quantities of ΔmixH, ΔmixS, and ΔmixG on u, x1, and x2 when the vapor pressures of the two components in a binary solution behave as

€

Pa = xaPa*euxb

2 / RT

We can begin with the definition of the free energy of a solution as

€

G = n1µ1 + n2µ2

= n1 µ1* + RT ln

P1

P1*

"

# $ $

%

& ' ' + n2 µ2

* + RT lnP2

P2*

"

# $ $

%

& ' '

= n1 µ1* + RT ln x1eux2

2 / RT" # $

% & ' + n2 µ2

* + RT ln x2eux12 / RT"

# $

% & '

= n1µ1* + n1RT ln x1 + n1ux2

2 + n2µ2* + n2RT ln x2 + n2ux1

2

Dividing both sides by the total number of moles n1 + n2 provides

€

G = x1µ1* + x1RT ln x1 + x1ux2

2 + x2µ2* + x2RT ln x2 + x2ux1

2

⇒ G − x1µ1* + x2µ2

*( ) = x1RT ln x1 + x1ux22 + x2RT ln x2 + x2ux1

2

⇒ ΔmixG = RT x1 ln x1 + x2 ln x2( ) + ux1x2 x1 + x2( )

Noting that x1 + x2 = 1, and dividing both sides by u, we have

€

ΔmixG u

=RTu

x1 ln x1 + x2 ln x2( ) + x1x2

Note the correct behavior that if either x1 or x2 is zero, there is no mixing, and there is no free energy of mixing. To determine the molar entropy of mixing, we can differentiate G with respect to T to obtain

€

ΔmixS u

= −∂∂T

ΔmixG u

%

& '

(

) *

= −Ru

x1 ln x1 + x2 ln x2( ) + 0

⇒ ΔmixS = −R x1 ln x1 + x2 ln x2( )

A good sanity check is to ensure that this quantity is always non-negative, and equal to zero for either x1 or x2 = 0, and one can see that this is satisfied. Finally, the molar enthalpy of mixing may be computed from H = G + TS or

€

ΔmixH u

=ΔmixG

u+

TΔmixS u

=RTu

x1 ln x1 + x2 ln x2( ) + x1x2 −TRu

x1 ln x1 + x2 ln x2( )$

% &

'

( )

= x1x2

which, interestingly, is always positive if u is positive and always negative if u is negative.

7. McQuarrie and Simon, 10-41. Paraphrase: Plot the behavior of the molar mixing free energy vs. x1 in the prior problem as a function of RT/u. Determine conditions where the curvature of the mixing free energy plots can be negative.

For the values of RT/u given (0.60, 0.50, 0.45, 0.40, and 0.35) it is a straightforward spreadsheet exercise to generate the necessary plots (in the plot, w is used as a variable name instead of u).

To evaluate regions of negative curvature, we need the second derivative of the mixing free energy as a function of x1. First derivative first…

€

∂∂x1

ΔmixG u

$

% &

'

( ) =

∂∂x1

RTu

x1 ln x1 + x2 ln x2( ) + x1x2*

+ ,

-

. /

=∂∂x1

RTu

x1 ln x1 + 1− x1( ) ln 1− x1( )[ ] + x1 1− x1( )1 2 3

4 5 6

=RTu

ln x1 +1− ln 1− x1( ) −1[ ] +1− 2x1

Second derivative next

€

∂2

∂x12ΔmixG

u

$

% &

'

( ) =

∂∂x1

RTu

ln x1 +1− ln 1− x1( ) −1[ ] +1− 2x1+ , -

. / 0

=RTu

1x1

+1

1− x1

$

% &

'

( ) − 2

=RTu

1x1 1− x1( )1

2 3 3

4

5 6 6 − 2

and the question is, when is the r.h.s. negative? This is equivalent to asking when is the first term on the r.h.s. less than 2. To answer that question, it is helpful to ask what is the minimum value that the quantity in brackets can take on in the range 0 < x1 < 1. To address that, we evaluate the first derivative

€

∂∂x1

1x1 1− x1( )$

% & &

'

( ) )

=∂∂x1

x1−1 1− x1( )−1$

% & ' ( )

= −x1−2 1− x1( )−1

+ x1−1 1− x1( )−2

=−(1− x1) + x1

x12 1− x1( )2

=2x1 −1

x12 1− x1( )2

It is evident that the only critical point is x1 = 1/2 (the numerator is zero). Simple consideration of the endpoints makes it clear that this critical point is indeed a minimum. And, plugging 1/2 into the original expression we see that its value is simply 4. So, if RT/u is less than 0.5, it is possible for the product of RT/u and the quantity in brackets to be less than 2.

8. McQuarrie and Simon, 10-49. Paraphrase: What are the solvent and solute activities and activity coefficients for a component having 25% mole fraction and obeying the vapor pressure equation P1 = 78.8x1exp(0.65x2

2+0.18x23)?

We may evaluate the vapor pressure of pure component 1 by setting x1 = 1 and x2 = 0, which gives 78.8. Within the solvent standard state (Raoult’s law), we have

€

a1 =P1P *

=78.8x1e

0.65x22+0.18x2

3( )78.8

= x1e0.65x2

2+0.18x23( )

= 0.25e0.65 0.75( )2

+0.18 0.75( )3" # $

% & '

= 0.39

And, as the activity coefficient γ is equal to the activity divided by the mole fraction, it is 0.39 / 0.25 = 1.6. In the solute standard state (Henry’s law),

€

a1 =P1

kH,1

We’re not given the Henry’s constant, but it is defined as the slope of P1 as x1 goes to zero. To find the slope, we should take the derivative of P1 with respect to x1

€

kH,1 = limx1→0

dP1dx1

= limx1→0

ddx1

78.8x1e0.65x2

2+0.18x23( )

= limx1→0

ddx1

78.8x1e0.65 1−x1( )2

+0.18 1−x1( )3$ % &

' ( )

= 78.8 limx1→0

e0.65 1−x1( )2

+0.18 1−x1( )3$ % &

' ( ) + x1

ddx1

e0.65 1−x1( )2

+0.18 1−x1( )3$ % &

' ( )

* + ,

- ,

. / ,

0 ,

= 78.8e 0.65+0.18[ ] + 0 × something finite( )= 78.8e 0.65+0.18[ ] =180.7

So, we can now evaluate

€

a1 =P1

kH,1=

78.8x1e0.65x2

2+0.18x23( )

180.7

=78.8 × 0.25e

0.65 0.75( )2+0.18 0.75( )3#

$ % & ' (

180.7= 0.17

and the Henry’s Law standard-state activity coefficient is 0.17 / 0.25 = 0.68.